# Variance and Standard Deviation of multiple dice rolls

I'm trying to determine what the variance of rolling $5$ pairs of two dice are when the sums of all $5$ pairs are added up (i.e. ranging from $10$ to $60$).

My first question is, when I calculate the variance using $E[X^2]-E[X]^2$ I get $2.91$, but my Excel spreadsheet and other sites I've googled give $3.5$ with no explanation of what me taking place. Which one is correct?

Second, to calculate the variance of a random variable representing the sum of the $5$ pairs (i.e. between $10$ and $60$), is it simply $5 \times Var(X)$? What about the standard deviation, is it $\sigma \sqrt{n}$?

Last, is there any difference between calculating the dice sums as "$5$ pairs of $2$ dice" and "$10$ dice"? Will it make a practical difference? (I find it easier to calculate it as $10$ dice).

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It is not the variance, but the expected value of a dice roll, $E(X)$ that is 3.5. – wnvl Nov 14 '12 at 1:53
The assumptions in the second and third part of the question are all correct. – wnvl Nov 14 '12 at 1:55
So, in other words the standard deviation of 5 pairs of 2 dice and the standard deviation of 10 dice is 5.3759? Can you confirm? – CodyBugstein Nov 14 '12 at 2:02

In your problem, there are five independent experiments, each of which is the sum of two die rolls. This is different from ten dice rolls. For example, you would expect a mean of 7 from your experiment, and 3.5 from the single dice rolls.

In excel, create two columns of five rows of random die rolls (=INT(RAND()*6)+1 in cells A1..B5), and then add the first two columns in the third column to make the random variable you want statistics on (=A1+B1, etc. in cells C1..C5).

After this, the excel built-in functions AVERAGE(C1:C5), VAR(C1:C5), and STDEV(C1:C5) can be used to compute the average ${\tt AVERAGE}=\frac{1}{N} \sum X_i$, sample variance ${\tt VAR}=\frac{1}{N-1} \sum (X_i-\bar{X})^2$, and sample standard deviation ${\tt STDEV} = \sqrt{{\tt VAR}}$.

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So we are tossing $10$ dice. Let $X_i$ be the result of the $i$-th toss. Let $Y=X_1+X_2+\cdots +X_{10}$. It seems that you want the variance of $Y$.

The variance of a sum of independent random variables is the sum of the variances. Now calculate the variance of $X_i$. This as usual is $E(X_i^2)-(E(X_i))^2$.

We know that $E(X_i)=3.5$. For $E(X_i^2)$, note that this is $$\frac{1}{6}(1^2+2^2+3^2+4^2+5^2+6^2).$$

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